Question

Six dogs pull a two-person sled with a total mass of 220 kg. The coefficient of kinetic friction between the sled and the snow is 0.08. The sled accelerates from rest at 0.75 m/s2 until it reaches a speed of 12 km/h. a) What is the work done by the dogs in the accelerating phase? b) What is the maximum power output by the dogs? c) What is the pulling force from the dogs (assumed to be horizontal)?

Answer 1

Answer:

a) 2457J

b) 558W

c) 337N

Explanation:

Assuming dogs started from rest.

$v=a*t\\t=\frac{a}{v}\\v=12\frac{km}{h}\frac{1000m}{km}*\frac{1h}{3600s}\\\\v=3.3m/s\\\\t=\frac{3.3m/s}{0.75m/s^2}\\t=4.4s$

and the displacement is given by:

$d=\frac{1}{2}*a*t^2\\d=7.3m$

Using the energy conservation formula:

$K_i+U_i+W_d+W_f=K_f+U_f$

Because the motion started from rest the initial kinetic energy is zero, the motion occurred in-ground level so the gravitational energy is zero too.

the work done by the friction force is given by:

$W_f=F_f*d*cos(\theta)\\W_f=\µ*m*g*d*cos(180)\\W_f=0.08*220kg*9.8m/s^2*7.3m*(-1)\\W_f=-1259J$

so:

$W_d=\frac{1}{2}*220kg*(3.3m/s)^2+1259J\\W_d=2457J$

The power is given by:

$P=\frac{W}{t}\\\\P=\frac{2457J}{4.4s}\\\\P=558W$

and the force exerted by the dogs:

$W_d=F_d*d*cos(\theta)\\F_d=\frac{W_d}{d*cos(0)}\\\\F=\frac{2457J}{7.3m*(1)}\\\\F=337N$